[Math] Examples of “Unusual” Classifications

big-listexamplesgm.general-mathematicssoft-question

When one says "classification" in math, usually one of a handful of examples springs to mind:

-Classification of Finite Simple Groups with 18 infinite families and 26 sporadic examples (assuming one believes the Classification is indeed complete)

-Classification of finte-dimensional semisimple Lie algebras with 4 infinite families and 5 exceptional examples

-Classification of (Simple, Formally Real) Jordan Algebras with 4 infinite families and 1 exceptional example

I'm sure there are other examples that non-algebraists would think of before these. All the examples I cited take the basic form of having several infinite families and some number of exceptional examples which do not fall into any of these families. Thus I am wondering:

Question: Does anyone know of examples of classifications of some mathematical objects such that the classification consists (A) only of infinite families or (B) only of a finite number of examples/ an infinite number of examples which do not seem to be closely related to one another (i.e. they do not "appear" to form any infinite families).

One example of case (B) that comes to mind would be finite dimensional division algebras over $\mathbb{R}$ of which there are 4. On the other hand, for case (B) I would like to rule out way too specific "classifications" such as "finite simple groups with an involution centralizer of such and such a form" since this is really a subclassification within the classification of FSG's. Although I am an algebraist, I would like to hear about examples from any branch of math, for comparison's sake.

(If anyone thinks of better tags for this, feel free to add or suggest them).

Best Answer

This feels like a somewhat silly example, but what about the classification of two (real) dimensional manifolds? They are two families of these, the orientable and non-orientable families classified by their genus/Euler characteristic.

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[Math] Does anyone want a pretty Maass form

[these are comments, not an answer, but there were too many for the comments box]

Hey---I wrote that code too! I did it to teach myself "practical Maass forms". I wrote in pari, not sage. I didn't do the example you did. Here's what I did, for what it's worth. First I tried a dihedral example. I used the Hilbert class field of $\mathbf{Q}(\sqrt{145})$; the class group is cyclic of order 4, giving a $D_8$ extension of $\mathbf{Q}$ with a faithful 2-dimensional representation. It's an easy exercise in factoring polynomials mod $p$ to compute traces of Frobenius, and I got the Hecke eigenvalues with little trouble.

But here's the big question: how do you know you got them right? Here's how I did it. I computed the first 200,000 Hecke eigenvalues, created the formal power series defining a function on the upper half plane as per usual, and then I evaluated it to many decimal places at lots of random points $z$ and $\gamma.z$ with $\gamma$ in the level. In all the cases I tried, the answers were the same to within experimental error. I concluded that probably I'd got everything working.

I also did an $S_3$ extension (the class group of $\mathbf{Q}(\sqrt{79})$) and a non-algebraic example coming from a Grossencharacter---this one had conductor 8 but eigenvalue not $1/4$.

I also tried to do an $A_5$ example! As you probably know from the $A_4$ example, an issue that needs resolving here is that the image of Galois in $GL_2(\mathbf{C}$) isn't $A_4$, it's the central extension, so you need to know how primes split in that last extension, which is computationally more expensive. Bjorn Poonen showed me a wonderful trick though, so I could do it. I computed $a_n$ for hundreds and thousands of $n$, built the function on the upper half plane, and checked to see if it was invariant by the level group. It wasn't :-( I concluded that either the Langlands program was wrong or my code was wrong, and I had a good idea which. [EDIT May 2012: for what it's worth I did actually get the code working in the end (in April 2011 in fact) -- my code was wrong (stupid error: all the "hard" code was fine but I had miscalculated $a_{1951}$!), and Langlands' programme still looks fine. AFAIK one still cannot prove that the candidate Maass form whose power series expansion I can compute a very long way is actually a Maass form.]

Here is the pari script for the 145 example, by the way:

N=200000;
f=x^4 - x^3 - 3*x^2 + x + 1;
ap(p)=if(p==5,-1,if(p==29,-1,if(issquare(Mod(p,5))&&issquare(Mod(p,29)),2*(matsize(factormod(f,p))[1]-3),0)));
chi(n)=kronecker(n,145);
v=vector(N,i,0);
v[1]=1;
for(i=2,N,fac=factor(i);k=matsize(fac)[1];\
 if(k>1,v[i]=prod(j=1,k,v[fac[j,1]^fac[j,2]]),\
 if(fac[1,2]==1,v[i]=ap(i),\
 p=fac[1,1];e=fac[1,2];v[i]=v[p]*v[p^(e-1)]-chi(p)*v[p^(e-2)]))\
);
F(z)=local(x,y,M);x=real(z);y=imag(z);M=ceil(11/y);if(M>N,error("y too small."));sqrt(y)*sum(n=1,M,if(v[n]==0,0,v[n]*besselk(1e-30*I,2*Pi*n*y)*cos(2*Pi*n*x)))

That's it! It's pretty self-explanatory. ap(p) returns the coefficient $a_p$ of the form. chi is the character of the form. If you run this the computer will pause for a few seconds while it computes the first 200,000 coefficients of the Maass form. After that it will give you a function $F$ on the upper half plane, which is defined by a Fourier expansion, and the miracle will be that it will be $\Gamma_1(145)$-invariant. For example, after running the code above, you can try this:

gp > z=-0.007+0.08*I
%5 = -0.007000000000000000000000000000 + 0.08000000000000000000000000000*I
gp > F(z)
%6 = 0.2101332751524672135753981488 + 0.E-30*I
gp > F(z/(145*z+1))
%7 = 0.2101332751524672135753981489 + 0.E-30*I
gp > %6-%7
%8 = -5.67979851 E-29 + 0.E-30*I

What this says is that for $z$ a random element of the upper half plane such that $z$ and $\gamma.z$ both have imaginary part which is not too small (if the im part is too small you need more Fourier coeffts), where here $\gamma=(1,0;145,1)$, $F$ evaluates, to within experimental error, to the same value at $z$ and $\gamma.z$.

Note that Marty's example is of $A_4$ type, so more interesting than this example, but a theorem of Langlands tells us that Marty's example really will be a cusp form.

Finite Simple Groups – Heuristic Argument for Classifiability

It is unlikely that there is any easy reason why a classification is possible, unless someone comes up with a completely new way to classify groups. One problem, as least with the current methods of classification via centralizers of involutions, is that every simple group has to be tested to see if it leads to new simple groups containing it in the centralizer of an involution. For example, when the baby monster was discovered, it had a double cover, which was a potential centralizer of an involution in a larger simple group, which turned out to be the monster. The monster happens to have no double cover so the process stopped there, but without checking every finite simple group there seems no obvious reason why one cannot have an infinite chain of larger and larger sporadic groups, each of which has a double cover that is a centralizer of an involution in the next one. Because of this problem (among others), it was unclear until quite late in the classification whether there would be a finite or infinite number of sporadics.

Any easy way to get around this has been overlooked by about a hundred finite group theorists.

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[Math] Does anyone want a pretty Maass form

Finite Simple Groups – Heuristic Argument for Classifiability

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